Help me "see" how to factor this dinomial

[AcousticBruce]

Hello folks. I pulled out my algebra and trigonometry book that I kept from college (that I never ended up going). I am brushing up with algebra right now and this is something that stumped me. If you do not mind, I would love to learn how I can think differently in order to complete this problem. Thank you.

$a^6-b^6$

I immediately see that I can do a difference of cubes like this...

$(a^2)^3-(b^2)^3$

and this leaves...

$(a^2-b^2)(a^4+a^2b^2 + b^4 )$

I am left with a difference of squares so I expand farther...

$(a-b)(a+b)(a^4 + a^2b^2 + b^4)$

I thought I was done, though something did look fishy about the trinomial and here is what the solution manual has...

$(a-b)(a+b)(a^2 - a*b + b^2)(a^2 + a*b + b^2)$

I was completely confused by the answer so I factored the last trinomial with a ti-89 and sure enough, it expanded farther.

Why does $(a^4+a^2b^2 + b^4 )$ factor into $(a^2 - a*b + b^2)(a^2 + a*b + b^2)$

How can I look at this and understand why?

 

[willem2]

Why does (a4+a2b2+b4)(a^4+a^2b^2 + b^4 ) factor into (a2−a∗b+b2)(a4+a∗b+b2)

it's a difference of squares (a^2 + b^2)^2 - (ab)^2

 

[AcousticBruce]

it's a difference of squares (a^2 + b^2)^2 - (ab)^2

I am sorry. I have no idea what you mean.

How is $(a^4 + a^2y^2 + b^4)$ a difference of squares?

 

[tommyxu3]

My opinion:
$a^4+a^2b^2+b^4=(a^4+2a^2b^2+b^4)-a^2b^2=(a^2+b^2)^2-(ab)^2=(a^2+b^2+ab)(a^2+b^2-ab),$ which is the result what you want.

 

[Tobias Funke]

After a while, it becomes pretty natural when seeing something like
$a^4+a^2b^2+b^4$
to notice that it looks awfully close to
$(a^2+b^2)^2$.

That's not the factorization of course, but it's only off by the single term $a^2b^2$, which is where tommyxu3 starts. So what I'm trying to say is that adding and subtracting $a^2b^2$ isn't some particularly clever trick---it's practically forced on you.

 

[micromass]

Here's a trick for factoring so-called homogeneous polynomials that works all the time. So let us start with a polynomial

$$a^3 - 6 a^2 b + 11ab^2 -6b^3$$

Homogeneous means that each term has the same degree. In this case, each term has degree $3$. These polynomials can always be factored as follows: factor out $b^3$ to get:

$$b^3 \left( \left(\frac{a}{b}\right)^3 - 6 \left(\frac{b}{a}\right)^2 +11\left( \frac{b}{a} \right)- 6\right)$$

Now we have an ordinary polynomial in $a/b$. So we need to factor $x^3 - 6x^2 + 11x - 6$. This has three roots: $1$, $2$ and $3$. So we can write it as

$$b^3\left(\frac{a}{b} - 1\right)\left(\frac{a}{b}- 2\right)\left(\frac{a}{b} - 3\right)$$

Factoring $b$ inside again, we get the desired factorization

$$(a - b)(a- 2b)(a- 3b)$$

 

[AcousticBruce]

My opinion:
$a^4+a^2b^2+b^4=(a^4+2a^2b^2+b^4)-a^2b^2=(a^2+b^2)^2-(ab)^2=(a^2+b^2+ab)(a^2+b^2-ab),$ which is the result what you want.

It took me a while to see it, but now I see it. Geez, this is crazy. I might not have ever figured this out.

After a while, it becomes pretty natural when seeing something like
$a^4+a^2b^2+b^4$
to notice that it looks awfully close to
$(a^2+b^2)^2$.

That's not the factorization of course, but it's only off by the single term $a^2b^2$, which is where tommyxu3 starts. So what I'm trying to say is that adding and subtracting $a^2b^2$ isn't some particularly clever trick---it's practically forced on you.

So when you say "forced" you mean this is not normal? How is subtracting $a^2b^2$ not clever? This reminds me of completing the squares, but it is so different then the problems I am used to.

 

[Mentallic]

A quicker way to arrive at the solution would be to factorize starting with a difference of squares, because you'll end up with two factors, each being a difference and sum of cubes, and both of those can be factored, so you'll arrive at the 4 factors you'll need almost straight away.

$$a^6-b^6$$
$$=(a^3)^2-(b^3)^2$$
$$=(a^3-b^3)(a^3+b^3)$$
$$=(a-b)(a^2+ab+b^2)(a+b)(a^2-ab+b^2)$$

 

[AcousticBruce]

Here's a trick for factoring so-called homogeneous polynomials that works all the time. So let us start with a polynomial

$$a^3 - 6 a^2 b + 11ab^2 -6b^3$$

Homogeneous means that each term has the same degree. In this case, each term has degree $3$. These polynomials can always be factored as follows: factor out $b^3$ to get:

$$b^3 \left( \left(\frac{a}{b}\right)^3 - 6 \left(\frac{b}{a}\right)^2 +11\left( \frac{b}{a} \right)- 6\right)$$

Now we have an ordinary polynomial in $a/b$. So we need to factor $x^3 - 6x^2 + 11x - 6$. This has three roots: $1$, $2$ and $3$. So we can write it as

$$b^3\left(\frac{a}{b} - 1\right)\left(\frac{a}{b}- 2\right)\left(\frac{a}{b} - 3\right)$$

Factoring $b$ inside again, we get the desired factorization

$$(a - b)(a- 2b)(a- 3b)$$

I am still studying this and I will sleep on it and delve into it excitedly tomorrow. I am loving this. Thank you so much.

A quicker way to arrive at the solution would be to factorize starting with a difference of squares, because you'll end up with two factors, each being a difference and sum of cubes, and both of those can be factored, so you'll arrive at the 4 factors you'll need almost straight away.

$$a^6-b^6$$
$$=(a^3)^2-(b^3)^2$$
$$=(a^3-b^3)(a^3+b^3)$$
$$=(a-b)(a^2+ab+b^2)(a+b)(a^2-ab+b^2)$$

I agree. This is the much easier route. Instead of going for a cube problem first, it could have been easily the difference of squares, this would have left me 2 separate cube problems to factor.

Though, if I am ever hit with $a^4+a^2b^2+b^4$ again, I need to have it much more together. Thank you for your input.

 

[Mentallic]

Though, if I am ever hit with $a^4+a^2b^2+b^4$ again, I need to have it much more together. Thank you for your input.

You should become familiar with the complete square in this form $x^2+y^2=(x+y)^2-2xy$ because it's often useful.

 

[micromass]

You should become familiar with the complete square in this form $x^2+y^2=(x+y)^2-2xy$ because it's often useful.

I don't think I've ever used this in my entire math career.

 

[AcousticBruce]

$$b^3 \left( \left(\frac{a}{b}\right)^3 - 6 \left(\frac{b}{a}\right)^2 +11\left( \frac{b}{a} \right)- 6\right)$$
I am going to assume you meant $a/b$ instead of $b/a$.Now we have an ordinary polynomial in $a/b$. So we need to factor $x^3 - 6x^2 + 11x - 6$. This has three roots: $1$, $2$ and $3$

$$b^3\left(\frac{a}{b} - 1\right)\left(\frac{a}{b}- 2\right)\left(\frac{a}{b} - 3\right)$$

$$(a - b)(a- 2b)(a- 3b)$$

I am completely confused by the -1 -2 and the -3. Can you help me understand this Micromass?

 

[tommyxu3]

I am completely confused by the -1 -2 and the -3. Can you help me understand this Micromass?

This is based on your knowing that $x^3-6x^2+11x-6$ has three root $x=1,2,3,$ and how to get it you have lots of methods, so $x^3-6x^2+11x-6=(x-1)(x-2)(x-3).$
If it's known, then for $(\frac{a}{b})^3-6(\frac{a}{b})^2+11(\frac{a}{b})-6$ we can get $(\frac{a}{b})^3-6(\frac{a}{b})^2+11(\frac{a}{b})-6=((\frac{a}{b})-1)((\frac{a}{b})-2)((\frac{a}{b})-3),$ so $a^3- 6a^2b+11ab^2-6b^3=b^3((\frac{a}{b})^3-6(\frac{a}{b})^2+11(\frac{a}{b})-6)=b^3((\frac{a}{b})-1)((\frac{a}{b})-2)((\frac{a}{b})-3)=(a-b)(a-2b)(1-3b).$

 

[tommyxu3]

I don't think I've ever used this in my entire math career.

It may be true for factoring is not a necessary skill for studying in advanced mathematics, but when learning the basic mathematics, it may be a helpful for one to get familiar some problems.

 

[AcousticBruce]

...
If it's known, then for $(\frac{a}{b})^3-6(\frac{a}{b})^2+11(\frac{a}{b})-6$ we can get $(\frac{a}{b})^3-6(\frac{a}{b})^2+11(\frac{a}{b})-6=((\frac{a}{b})-1)((\frac{a}{b})-2)((\frac{a}{b})-3),$
...

This is exactly what I do not understand. Wher does the -1 -2 and -3 come from. Thank you

 

[TeethWhitener]

I don't think I've ever used this in my entire math career.

It (kind of) shows up in my favorite proof of the Pythagorean theorem: http://www.mathsisfun.com/geometry/pythagorean-theorem-proof.html

 

[SteamKing]

This is exactly what I do not understand. Wher does the -1 -2 and -3 come from. Thank you

You have a polynomial in terms of (a/b): namely f (a/b) = (a/b)³ - 6(a/b)² + 11(a/b) - 6

We can clean this expression up a bit by letting (a/b) = x.

Now, the polynomial can be written: f(x) = x³ - 6x² + 11x - 6

This is a polynomial of degree three, which means it has three roots.

By setting f(x) = 0, one can find the three roots, i.e. values of x, which make this equation true.

In other words, find the three values of x such that x³ - 6x² + 11x - 6 = 0

There are several ways these values of x can be found; trial and error is one way.

If we assume x = 1, then 1 - 6 + 11 - 6 = 0. We've found a root.

This also means that (x - 1) is a factor of the original polynomial. We can use polynomial long division to divide x³ - 6x² + 11x - 6 by (x - 1) and find a quadratic quotient. But this seems a bit messy right now. Let's continue guessing roots.

If we assume x = 2, then 8 - 24 + 22 - 6 = 0. We've found another root. (x - 2) is also a factor of the original polynomial. There's one root left to find.

If we assume x = 3, then 27 - 54 + 33 - 6 = 0. We've found the third root. (x - 3) is the third and final factor of the original polynomial.

Thus, we can say that x³ - 6x² + 11x - 6 = (x - 1)(x - 2)(x - 3)

 

[AcousticBruce]

...
In other words, find the three values of x such that x³ - 6x² + 11x - 6 = 0

There are several ways these values of x can be found; trial and error is one way.

If we assume x = 1, then 1 - 6 + 11 - 6 = 0. We've found a root.

This also means that (x - 1) is a factor of the original polynomial. We can use polynomial long division to divide x³ - 6x² + 11x - 6 by (x - 1) and find a quadratic quotient. But this seems a bit messy right now. Let's continue guessing roots.

If we assume x = 2, then 8 - 24 + 22 - 6 = 0. We've found another root. (x - 2) is also a factor of the original polynomial. There's one root left to find.

If we assume x = 3, then 27 - 54 + 33 - 6 = 0. We've found the third root. (x - 3) is the third and final factor of the original polynomial.

Thus, we can say that x³ - 6x² + 11x - 6 = (x - 1)(x - 2)(x - 3)

I really appreciate all of the effort put into explaining this too me!

Would you are anyone else here mind giving me another problem that works out similar to this one so I can practice?
$$a^3 - 6 a^2 b + 11ab^2 -6b^3$$

 

[micromass]

$a^2 - 6ab + 8b^2$.

 

[AcousticBruce]

$a^2 - 6ab + 8b^2$.

$a^2 - 6ab + 8b^2$
this factors pretty easily to
$(a-2b)(a-4b)$

But that is not what I was wanting to learn... so Ill start again with the roots method.

$a^2 - 6ab + 8b^2$3

I will factor out $b^2$

$b^2((\frac{a}{b})^2-6(\frac{a}{b})+8)$

make it simple with

$x^2-6x+8$

this factors into $(x-2)(x-4)$ which gives the two roots 2 and 4.
$f(2) = 0$ (2)^2-6(2)+8 = 0
$f(4) = 0$ (4)^2-6(4)+8 = 0

so we get

$b^2(\frac{a}{b}-2)(\frac{a}{b}-4)$

$b^2$ multiplied...

$(a-2b)(a-4b)$I really appreciate learning this new way to factor. It really opened up my eyes with polynomial algebra.

 

[micromass]

Do you know complex numbers? If so, try to factor your original problem $a^4 + a^2 b^2 + b^4$ in four pieces instead of two.

 

[AcousticBruce]

Do you know complex numbers? If so, try to factor your original problem $a^4 + a^2 b^2 + b^4$ in four pieces instead of two.

Well, I know that $i^2 = -1$ and $i^4 = (-1)(-1) = 1$. I am just unaware how to make the polynomial equal zero.

$a^4 + a^2 b^2 + b^4$

factor out $b^4$

$b^4(\frac{a}{b})^4+(\frac{a}{b})^2+1)$

convert $\frac{a}{b}$ to $x$

$x^4+x^2+1$

Hopefully you do not have to give me the answer here, but how should I think about this based off of my knowledge of imaginary numbers?

 

[micromass]

This polynomial can be handed perfectly by setting $y=x^2$. You'll get a quadratic polynomial $y^2 + y + 1$. This is a really useful trick.

 

[AcousticBruce]

This polynomial can be handed perfectly by setting $y=x^2$. You'll get a quadratic polynomial $y^2 + y + 1$. This is a really useful trick.

Man.. I need another hint. I used the quadratic formula with $y^2 + y + 1$ and got $-1-i$ and $-1+i$.

Am I at least going in the right direction?

 

[micromass]

Those are not the correct roots. Can you show how you found them?

 

[AcousticBruce]

Those are not the correct roots. Can you show how you found them?

That is because I did it wrong. :/

${\frac{{ -(1) \pm \sqrt {(1)^2 - 4(1)(1)} }}{{2(1)}}}$

I believe these are the answers.

$-\frac{1}{2} + \frac{\sqrt{3}}{2}\cdot i$

$-\frac{1}{2} - \frac{\sqrt{3}}{2}\cdot i$Edit:: Is there another way to figure this out without the quadratic formula?

 

[micromass]

That is because I did it wrong. :/

${\frac{{ -(1) \pm \sqrt {(1)^2 - 4(1)(1)} }}{{2(1)}}}$

I believe these are the answers.

$-\frac{1}{2} + \frac{\sqrt{3}}{2}\cdot i$

$-\frac{1}{2} - \frac{\sqrt{3}}{2}\cdot i$

Ok, so you have found two possible $y$ values. Now you know that $x^2 = y$, so this will get you (in general) four possible $x$ values $A$, $B$, $C$, $D$. Then you can factorize your polynomial as $(x- A)(x-B)(x-C)(x-D)$.

Edit:: Is there another way to figure this out without the quadratic formula?

Yes sure, but they're not easier.

 

[tommyxu3]

Ok, so you have found two possible $y$ values. Now you know that $x^2 = y$, so this will get you (in general) four possible $x$ values $A$, $B$, $C$, $D$. Then you can factorize your polynomial as $(x- A)(x-B)(x-C)(x-D)$.

But most factorizing requires to get the coefficients in $\mathbb{R},$ or the final factorizing is just another present of finding out the roots of the equation.

 

[micromass]

But most factorizing requires to get the coefficients in $\mathbb{R},$ or the final factorizing is just another present of finding out the roots of the equation.

Even then, factoring in $\mathbb{C}$ is generally easier and can give you a factorization in $\mathbb{R}$ in a very straightforward way. And yes, factoring is pretty close to finding roots of the equation.